Đáp án:
$\lim\dfrac{1}{\sqrt{n^2+2} -\sqrt{n^2+4}} =-\infty$
Giải thích các bước giải:
$\quad \lim\dfrac{1}{\sqrt{n^2+2} -\sqrt{n^2+4}}$
$=\lim\dfrac{\sqrt{n^2+2} +\sqrt{n^2+4}}{(\sqrt{n^2+2} -\sqrt{n^2+4})(\sqrt{n^2+2} +\sqrt{n^2+4})}$
$=\lim\dfrac{\sqrt{n^2+2} +\sqrt{n^2+4}}{-2}$
$=-\dfrac12\lim(\sqrt{n^2+2} +\sqrt{n^2+4})$
$= -\dfrac12\lim\left[n\left(\sqrt{1+\dfrac{2}{n^2}} +\sqrt{1+\dfrac{4}{n^2}}\right)\right]$
$=-\dfrac12\cdot +\infty\cdot (\sqrt{1+0} +\sqrt{1+0})$
$= -\infty$
